Chapter 8, Problem 67A

(a)

To determine

The resulting angular velocity of the top.

Answer to Problem 67A

The resultant angular velocity of the top is $750\text{rad/s}$ .

Explanation of Solution

Given:

The initial linear velocity of the string is $0\text{m/s}$ .

The final linear velocity of the string is $v=3.0\text{m/s}$ .

The diameter of the rod is $d_{\text{rod}}=8.0\text{mm}$ .

The mass of the disk is $m=0.0125\text{kg}$ .

The diameter of the disk is $d_{\text{disk}}=3.5\text{cm}$ .

The time interval is $\Delta t=0.50\text{s}$ .

Formula used:

The expression for the angular velocity $\left(\omega \right)$ of a disk of radius $\left(r\right)$ as follows:

  $\omega =\frac{v}{r}$

Here, $v$ is the linear velocity of the disk.

Calculation:

The radius $\left(r\right)$ of the rod is,

  $\begin{array}{l}{r}_{\text{rod}}=\frac{d_{\text{rod}}}{2}\\ r_{\text{rod}}=\left\{\frac{8\text{mm}\times \left(\frac{1\text{m}}{1000\text{mm}}\right)}{2}\right\}\\ r_{\text{rod}}=0.004\text{m}\end{array}$

The resultant angular velocity of top is,

  $\begin{array}{l}\omega =\frac{v}{r_{\text{rod}}}\\ \omega =\frac{3.0\text{m/s}}{0.004\text{m}}\\ \omega =750\text{rad/s}\end{array}$

Conclusion:

Therefore, the resultant angular velocity of the top is $750\text{rad/s}$ .

(b)

To determine

The force exerted on the string.

Answer to Problem 67A

The force exerted on the string is $F=1.5\text{N}$ .

Explanation of Solution

Given:

The initial linear velocity of the string is $0\text{m/s}$ .

The final linear velocity of the string is $v=3.0\text{m/s}$ .

The diameter of the top is $d_{\text{rod}}=8.0\text{mm}$ .

The mass of the disk is $m=0.0125\text{kg}$ .

The diameter of the toy is $d_{\text{disk}}=3.5\text{cm}$ .

The time interval is $\Delta t=0.50\text{s}$ .

Formula used:

The expression for the angular acceleration $\left(\alpha \right)$ of the disk of radius $r$ is,

  $\alpha =\frac{Fr}{I}$

Here, $I$ is the moment of inertia of the system and $F$ is the force acting on the system.

Calculation:

Frompart (a), the resulting angular velocity of the top is $\Delta \omega =750\text{rad/s}$ .

The radius $\left(r_{\text{disk}}\right)$ of the disk is,

  $\begin{array}{l}{r}_{\text{disk}}=\frac{d_{\text{disk}}}{2}\\ r_{\text{disk}}=\left\{\frac{3.5\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)}{2}\right\}\\ r_{\text{disk}}=0.018\text{m}\end{array}$

The force $\left(F\right)$ exerted on the string as follows:

  $\begin{array}{l}F=\frac{I\alpha }{r_{\text{rod}}}\\ F=\frac{m{r}_{\text{disk}}^2}{r_{\text{rod}}}\times \left(\frac{\Delta \omega }{\Delta t}\right)\\ F=\frac{0.0125\text{kg}\times {\left(\text{0}\text{.018}\text{m}\right)}^2}{0.004\text{m}}\times \left(\frac{750\text{rad/s}}{0.50\text{s}}\right)\\ F=1.5\text{N}\end{array}$

Conclusion:

Therefore, the force exerted on the string is $F=1.5\text{N}$ .